Uniform convergence of $ \zeta'(x) = \sum \frac{\ln(k)}{k^x}$

Question

As a part of a proof that the $\zeta$ function is differentiable I want to check that the series: $$-\sum_{k=1}^\infty \frac{\ln(k)}{k^x}$$ Converges uniformly for $x \in (a, \infty)$ whenever $a>1$, I was thinking of using the Weierstrass M-test, but I got stuck.

This is what I have so far: We will first remark that: $$|f'_n|=\left|-\frac{\ln(n)}{n^x}\right|=\frac{\ln(n)}{n^x} \leq \frac{\ln(n)}{n^a} $$

We now notice since $n\geq 1$ that $\ln(n)<n $ and since $a>1$, we have $a \ln(n)= \ln(n^a) < n^a \implies \frac{1}{a}>\frac{\ln(n)}{n^a} $

But I don't know how to finish the M-test.

Answer 1

Uniform convergence on $[a,+\infty)$, where $a>1$. We will use the Weierstrass M-test.
For $x \in [a,+\infty)$ we have $$ \frac{\log(n)}{n^x} \le \frac{\log(n)}{n^a} $$

By the integral test${}^1$, the series $$ \sum_{n=1}^\infty \frac{\log(n)}{n^a} $$ converges, and therefore by the Weierstrass M-test, $$ \sum_{n=1}^\infty \frac{\log(n)}{n^x} $$ converges uniformly on $[a,+\infty)$.

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${}^1$If $a>1$, then $\log(x)/x^a$ is eventually decreasing, and $$ \int_1^\infty\frac{\log(x)}{x^a}\;dx = \frac{1}{(a-1)^2} $$

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added: Method for the integral.
Integrate by parts $$ \int_1^\infty \frac{\log(x)}{x^a}\;dx = \left. \frac{\log(x)}{x^{a-1}}\right|_{x=0}^{x=\infty} +\int_1^\infty\left(\frac{a\log(x)}{x^a}-\frac{1}{x^a}\right)dx $$

So, writing $S=\int_1^\infty\frac{\log(x)}{x^a}\;dx$ we get $$ S = aS - \frac{1}{a-1} $$ and the solution is $$ S = \frac{1}{(a-1)^2} $$

In fact, instead of using the integral test, we could more easily do a comparison: $\frac{\log(n)}{n^a} < \frac{1}{n^b}$ for large $n$, where $1<b<a$ so that $\sum \frac{1}{n^b} < \infty$.

Answer 2

To prove uniform convergence for $x \in [a,\infty)$ where $a > 1$ by a straightforward comparison, take $b$ such that $a-b > 1$.

We have

$$\frac{\log k}{k^x} \leqslant \frac{\log k}{k^ a} = \frac{1}{b}\frac{\log k^b}{k^a} \leqslant \frac{1}{b}\frac{k^b}{k^a} = \frac{1}{b} \frac{1}{k^{a-b}},$$

and $\displaystyle \sum \frac{1}{k^{a-b}}$ is a convergent p-series. Uniform convergence follows by the Weierstrass M -test.

The series fails to converge uniformly on the open interval $(1,\infty)$. Note that

$$\tag{*} \begin{align}\sup_{x \in (1,\infty)}\left|\sum_{k=n+1}^{\infty}\frac{\log k}{k^x}\right| &\geqslant \sup_{x \in (1,\infty)}\left|\sum_{k=n+1}^{2n}\frac{\log k}{k^x}\right|\\ &\geqslant \sup_{x \in (1,\infty)}n \frac{\log n}{(2n)^{x}} \\ & \geqslant n \frac{\log n}{(2n)^{1+1/n}} \\ &= \frac{\log n}{2 (2n)^{1/n}}\end{align}$$

Since $(2n)^{1/n} \to 1 $ and $\log n \to \infty$, the RHS of (*) does not converge to $0$ as $n \to \infty$.